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Inegalitatea 3, exponentiala, x^y+y^x \geq 1
Posted: Thu Sep 27, 2007 4:09 am
by Cezar Lupu
Fie \( x, y>0 \). Aratati ca \( x^y+y^x>1 \).
Posted: Sun Oct 14, 2007 9:55 pm
by Filip Chindea
Evident putem considera \( x, y < 1 \). Apoi substituim \( x = \frac{1}{1 + u} \), \( y = \frac{1}{1 + v} \), \( u, v > 0 \). Utilizand inegalitatea lui Bernoulli, deducem \( (1+u)^y < 1 + uy \) (deoarece \( y \in (0,1) \)). Deci \( x^y = \frac{1}{(1+u)^y} > \frac{1}{1 + uy} = \frac{1 + v}{1 + u + v} \), si analog \( y^x > \frac{1 + u}{1 + u + v} \). In concluzie, \( x^y + y^x > \frac{2 + u + v}{1 + u + v} = 1 + \frac{1}{1 + u + v} > 1 \).